Question: Someone must have beaten me here. Devora thinks, dismayed by the empty treasure chest. Then, just as she's about to close the lid, she notices a map (shown below). "Curse these trig-loving pirates!" Devora mutters as she looks at the map. Then, after thinking for a moment, she walks back to the entrance of the secret cave, measuring $48$ meters along the way. After performing a calculation, Devora walks back to the empty treasure, faces the entrance, and turns a certain number of degrees to her left before walking $89$ meters to the treasure. Supposing that the angle at the Cave entrance is acute and that Devora's calculations were correct, how many degrees did she turn? Do not round during your calculations. Round your final answer to the nearest degree.
Converting the problem into geometrical terms Our problem can be modeled by the following triangle $\triangle ABC$, where we want to find $\angle B=\theta$. $A$ $B$ $C$ $30^\circ$ $\theta$ $\alpha$ $89\,\text{m}$ $48\,\text{m}\;\;\;\;\;\;\;\;\;\;\;\;$ Since we are given two side lengths, we can use the law of sines to find $\angle A=\alpha$. Then, we will find $\theta$ using the fact that $\theta=180^\circ-30^\circ-\alpha$. When using the law of sines we have to keep in mind the ambiguous case, where the angle can be either acute or obtuse. In our case, we are explicitly given that $\alpha$ is acute. Using the law of sines $\begin{aligned} \dfrac{\sin(A)}{BC}&=\dfrac{\sin(C)}{AB}\\\\ \dfrac{\sin(\alpha)}{89} &= \dfrac{\sin(30^\circ)}{48} \gray{\text{Substitute}} \\\\ \sin(\alpha) &= \dfrac{89\cdot \sin(30^\circ) }{48} \\\\ \alpha&= \sin^{-1}\left(\dfrac{89\cdot \sin(30^\circ) }{48}\right) \\\\ \alpha&\approx 68^\circ \end{aligned}$ Now, $\theta=180^\circ-30^\circ-\alpha=82^\circ$. The answer Devora turned $82^\circ$ to the left.